- 41 名前：デフォルトの名無しさん mailto:sage [2008/06/09(月) 13:25:31 ]
- >>20
とりあえず a<b ならそれなりの解が出るソースコード
解法ｐｌｚ
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
int pattern1(double a, double b, double c){
int x, y;
y=(a-2*c)/(sqrt(3.)*c)+1;
x=b/(2*c);
return x*y-y/2;
}
int pattern2(double a, double b, double c){
int x, y;
y=a/(2*c);
x=b/(2*c);
return x*y;
}
int main(int argc, char *argv[]){
double a=20., b=30., c=3.;
int num, maximum;
switch(argc){
case 4: c=atof(argv[3]);
case 3: b=atof(argv[2]);
case 2: a=atof(argv[1]);
}
maximum=pattern1(a, b, c);
num=pattern2(a, b, c);
if(num>maximum) maximum=num;
printf("円の数:%d 使用率:%.3f%%\n", maximum, (c*c*M_PI*maximum)/(a*b)*100);
return 0;
}
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